This is because solids have less energy than liquids do, hence it takes more energy to excite a solid into its gaseous phase than it does a liquid.
When opposed to merely reducing their separation, from solid to liquid, the energy needed to totally separate the molecules as they move from liquid to gas is substantially higher. The latent heat of vaporization is therefore bigger than the latent heat of fusion for this reason.
<h3>
What is heat of sublimation?</h3>
The amount of energy required to change one mole of a substance from its solid to its gaseous state under particular conditions—typically the standard ones—is known as the enthalpy of sublimation or heat of sublimation (STP). A solid's worth is based on its cohesive energy.
<h3>
What is heat of vaporization?</h3>
The term "enthalpy of vaporization," which is often referred to as "heat of vaporization" or "heat of evaporation," refers to the amount of energy that must be applied to a liquid substance in order to cause a part of that substance to transform into a gas. Vaporization's enthalpy varies with the pressure at which the transition takes place.
Learn more about heat of sublimation: brainly.com/question/13200793
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The correct question is:
Why heat of the sublimation of a substance is greater than the heat of vaporization?
Answer:
Explanation:
<u>Molecular formula from Glucose:</u>
C₆H₁₂O₆
<u>3 moles of Glucose:</u>
3C₆H₁₂O₆
In 1 mole of Glucose, there are 12 hydrogen atoms.
<u>In 3 moles:</u>
= 12 × 3
= 36 H atoms
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Answer:
![[H^+]=5x10^{-13}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D5x10%5E%7B-13%7DM)
![[OH^-]=0.02M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.02M)
Explanation:
Hello there!
In this case, according to the given ionization of magnesium hydroxide, it is possible for us to set up the following reaction:
Thus, since the ionization occurs at an extent of 1/3, we can set up the following relationship:
![\frac{1}{3} =\frac{x}{[Mg(OH)_2]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%20%3D%5Cfrac%7Bx%7D%7B%5BMg%28OH%29_2%5D%7D)
Thus, x for this problem is:
![x=\frac{[Mg(OH)_2]}{3}=\frac{0.03M}{3}\\\\x= 0.01M](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%5BMg%28OH%29_2%5D%7D%7B3%7D%3D%5Cfrac%7B0.03M%7D%7B3%7D%5C%5C%5C%5Cx%3D%20%200.01M)
Now, according to an ICE table, we have that:
![[OH^-]=2x=2*0.01M=0.02M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D2x%3D2%2A0.01M%3D0.02M)
Therefore, we can calculate the H^+, pH and pOH now:
![[H^+]=\frac{1x10^{-14}}{0.02}=5x10^{-13}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5Cfrac%7B1x10%5E%7B-14%7D%7D%7B0.02%7D%3D5x10%5E%7B-13%7DM)
Best regards!
Yeah no one is gonna read all that babe.
Pretty sure it’s Mixture if I’m not wrong
